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\title{\vspace{-4cm}\textbf{河北师范大学数学分析真题}}
\author{宁鑫雨}
\date{\today}
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\begin{document}
\date{}
\section*{2013年数学分析}
\begin{problem}计算下列各题[本题满分30分，每小题10分]\\
    1、求极限$\lim_{x\to0}\frac{e^{x}-\sin x-1}{1-\sqrt{1-x^{2}}}$\\
    2、计算积分$I=\int_0^{2\pi}\frac{d\theta}{2+\cos\theta}.$\\
    3、计算$I=\iint_{\sum}(8y+1)xdydz+2(1-y^{2})dzdx-4yzdxdy$  其中$\sum$是曲面$y-1=x^{2}+z^{2}(1\leq y\leq3)$的外侧.
\end{problem}
\begin{solution}[本题30分,每题10分]\\
    1)\\
    $\begin{aligned}
        &\lim_{x\to0}\frac{e^{x}-\sin x-1}{1-\sqrt{1-x^{2}}}=\lim_{x\to0}\frac{e^{x}-\cos x}{-\frac{1}{2}(1-x^{2})^{-\frac{1}{2}}\cdot-2x} \\
        &=\lim_{x\to0}\frac{(e^{x}-\cos x)\sqrt{1-x^{2}}}{x}\\
        &=\lim_{x\to0}(e^{x}+\sin x)\sqrt{1-x^{2}}+(e^{x}-\cos x)\cdot\frac{x}{\sqrt{1-x^{2}}} \\
        &=1\\
    \end{aligned}$\\
    2)$I=\int_{0}^{2\pi}\frac{\ d \theta}{2+\cos\theta}$\\
    令$t=\tan\frac{\theta}{2}$\\
    $\therefore \cos \theta=\frac{1-t^{2}}{1+t^{2}}$\\
    $\theta=\arctan t,\therefore \ d \theta=\frac{2}{1+t^{2}}\ d t$\\
    $\begin{aligned}
        &I=\int_{0}^{2\pi}\frac{\ d \theta}{2+\cos\theta}=2\int_{0}^{+\infty}\frac{\frac{2}{1+t^{2}}}{2+\frac{1-t^{2}}{1+t^{2}}}\ d t=2\int_{0}^{+\infty}\frac{2}{3+t^{2}}\ d t \\
        &=4\int_{0}^{+\infty}\frac{1/3}{1+(\frac{t}{\sqrt3})^{2}}\ d t=4\sqrt{3}\int_{0}^{+\infty}\frac{\frac{1}{3}}{1+(\frac{t}{\sqrt3})^{2}}\ d (\frac{t}{\sqrt{3}}) \\
        &=\frac{4\sqrt{3}}{3}\arctan\frac{t}{\sqrt{3}}\vert_{0}^{+\infty}=\frac{2\sqrt{3}}{3}\pi 
        \end{aligned}$\\
    3)34$\pi$
\end{solution}


\begin{problem}[本题满分10分]\\
    1、设$\lim_{x\to0}f(x^{3})=A$,证明：$\lim_{x\to0}f(x){=}A$.\\
    2、设$\lim_{x\to0}f(x^{4})=A$,据理说明是否成立$\lim_{x\to0}f(x){=}A$？
\end{problem}
\begin{solution}[本题10分]\\
    证明:\\
    1)$\because\lim_{x\to0}f(x^{3})=A$\\
    $\therefore\forall \varepsilon>0,\exists\delta_{1}>0,$当$0<|\varphi|<\delta_{1}$时,$\vert f(x^{3})-A\vert<\varepsilon $\\
    取$\delta=\delta_{1}^{3}$,$\therefore 0<|x|<\delta$时,$\therefore 0<|\sqrt[3]{x}|<\sqrt[3]{\delta}=\delta_{1}$\\
    $\therefore|f(\sqrt[3]{x})^{3}-A|=|f(x)-A|<\varepsilon $\\
    $\therefore\lim_{x\to0}f(x)=A$\\
    2)不一定\\
    例:\\
    $f(x)=sgnx=\begin{cases}1&x>1\\0&x=1\\-1&x<1\end{cases}$\\
    $f(x^4)=\mathrm{~sgn~}x^4=\left\{\begin{array}{ll}0,&x=1\\1,&x\neq1\end{array}\right.$\\
    $\lim_{x\to0}f(x^{4})=1$,$\lim_{x\to0}f(x)$不存在\\
    $\therefore$结论不一定成立
\end{solution}


\begin{problem}[本题满分10分]\\
    设函数$f(x)$在$(a,b)$内连续，且$f(a+0)=f(b-0)=+\infty$,证明：$f(x)$在$(a,b)$内能取到最小值.
    \end{problem}    
\begin{solution}[本题10分]\\
   证明:\\
   取$M=\vert f(\frac{a+b}{2})\vert\geq0$\\
   由已知有$\lim_{x\rightarrow a^{+}}f(x)=+\infty$和$\lim_{x\to b^{-}}f(x)=+\infty $\\
   $\therefore \forall \varepsilon>0,\exists\delta=\frac{1}{3}(b-a)$,当$x\in(a,a+\delta)$时,有$f(x)>M$\\
   当$x\in(b-\delta，b)$时,$f(x)>M$\\
   $\therefore\forall x\in(a,a+\delta)\cup(b-\delta,b)$有$f(x)>M\geq f(\frac{a+b}{2})$\\
   在$x\in[a+\delta,b-\delta]$上,$f(x)$连续\\
   $\therefore\exists\xi\in[a+\delta,b-\delta],f(\xi)\leq f(x)$\\
   $\because\frac{a+b}{2}\in[a+\delta,b-\delta],\therefore f(\xi)\leq f(\frac{a+b}{2})$\\
   在$x\in(a,a+\delta)\cup(b-\delta,b),f(\frac{a+b}{2})\leq f(x)$\\
   $\therefore$有$f(\xi)\leq f(x)$成立,$x\in(a,a+\delta)\cup(b-\delta,b)$\\
   综上$f(x)$在$(a,b)$内能取得最小值$f(\xi),\xi \in [a+\delta,b-\delta]\subset(a,b)$\\
\end{solution}


\begin{problem}[本题满分10分]\\
    设$f(x)$是定义在$R$上的函数，并且对任意的$x_{1},x_{2}\in{R}$，都有$f(x_{1}+x_{2})=f(x_{1})f(x_{2})$,若$f^{\prime}(0)=1$,证明对任意的$x\in{R}$,都有$f^{\prime}(x)=f(x)$.
\end{problem}
\begin{solution}[本题10分]\\
   证明:\\
   令$x_{1}=x_{2}=0$\\
   $\therefore f(0)=f(0)^{2}$\\
   $\therefore f(0)=0或f(0)=1$\\
   若$f(0)=0$,则$f(0+x)=0\cdot f(x)=0$,即$f(x)=0$\\
   $\therefore f^{\prime}(x)=0,\therefore f^{\prime}(0)=0\neq1,$矛盾$\therefore f(0)=1$\\
   $\begin{aligned}
    &f^{\prime}(x)=\lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\\
    &=\lim_{\Delta x\to0}\frac{f(x)\cdot f(\Delta x)-f(x)}{\Delta x} \\
    &=\lim_{\Delta x\rightarrow0}f(x)\cdot\frac{f(\Delta x)-1}{\Delta x} \\
    &=f(x)\cdot\lim_{\Delta x\rightarrow0}\frac{f(\Delta x)-f(0)}{\Delta x-0}\\
    &=f(x)\cdot f^{\prime}(0)=f(x) 
    \end{aligned}$
\end{solution}
    

\begin{problem}[本题满分15分]\\
    证明对任意自然数$n$,方程
    $$
    x^{n}+.x^{n-1}+\cdots+x^{2}+x=1
    $$
    在区间$[0,1]$内总有唯一的实根$x_{n}$，并求$\lim_{n\to\infty}x_n$.
\end{problem}    
\begin{solution}[本题15分]\\
  证明:\\
  设$f(x)=x^{n}+x^{n-1}+\cdots+x^{2}+x-1$,且一定连续\\
  $f(0)=-1<0, f(1)=n-1\geq0$\\
  由介值定理,$\exists x_{n}\in(0,1)\subset [0,1]$有$f(x_{n})=0$\\
  又$f^{\prime}(x)=nx^{n-1}+(n-1)x^{n-2}+\cdots+1>0$,$\therefore f(x)$单调递增\\
  $\therefore f(x)$在$[0,1]$上总有唯一实根\\
  可知$x^{n}+x^{n-1}+\cdots+x^{2}+x=\frac{x(1-x^{n})}{1-x}=1$\\
  $\because x_{n}\in[0,1]$为唯一实根,$\therefore \frac{x_{n}(1-{x_{n}}^{n})}{1-x_{n}}=1$\\
  $\therefore x_{n}=\frac{1+{x_{n}}^{n+1}}{2}$\\
  $\therefore\lim_{x\to\infty}x_{n}=\lim_{n\to\infty}\frac{1+{x_{n}}^{n+1}}{2}=\frac{1}{2}$
\end{solution}
    

\begin{problem}[本题满分15分]\\
    设$S$为有界数集，证明：若$\inf S=a\notin S$,则存在严格递减数列$\{x_{n}\}\subset S$,使得$\lim_{n\to\infty}x_n={a}$.
\end{problem}
\begin{solution}[本题15分]\\
   证明:\\
   $\because a$为$S$的下确界,$\therefore \forall \varepsilon>0,\exists x\in S,$使得$x<a+\varepsilon$\\
   $\because a\in S,\therefore x>a,\therefore a<x<a+\varepsilon$\\
   取$\varepsilon_{1}=1,\exists x_{1}\in S,$使得$a<x_{1}<a+\varepsilon_{1}$\\
   取$\varepsilon_{2}=\min\{\frac{1}{2},x_{1}-a\},\exists x_{2}\in S$使得$a<x_{2}<a+\varepsilon_{2}$,\\
   且$x_{2}<a+\varepsilon_{2}<a+x_{1}-a=x_{1}$\\
   一般地，按上述步骤得到$x_{n-1}\in S$\\
   取$\varepsilon=\min{\frac{1}{n},x_{n-1}-a},\exists x_{n}\in S,$使得\\
   $a<x_n<a+\varepsilon_{n},$且有$x_{n}<x_{n-1}$\\
   上述过程无限进行下去，得到${x_{n}}\subset S$,为单调递减数列\\
   且$a-\varepsilon_{n}<a<a+\varepsilon_{n}$\\
   $\therefore |x_{n}-a|<\varepsilon_{n}$\\
   $\therefore \lim_{n\to\infty}x_{n}=a$
\end{solution}


\begin{problem}[本题满分15分]\\
    设${a_{1}=}\sqrt{2} .\quad a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{1}{a_{n}}\right)(n=1,2,\cdots)$.证明\\
    1、极限$\lim_{n\to\infty}a_n$存在.\\
    2、级数$\sum_{n=1}^{\infty}\left(\frac{a_{n}}{a_{n+1}}-1\right)$收敛.
\end{problem}
\begin{solution}[本题15分]
    证明:\\
    1)$a_{n+1}=\frac{1}{2}(a_{n}+\frac{1}{a_{n}})\geq\frac{1}{2}\cdot2\sqrt{1}=1$有下界\\
    $a_{n+1}-a_{n}=\frac{{a_{n}}^2+1}{2a_{n}}-a_{n}=\frac{a_{n}^{2}+1-2a_{n}^{2}}{2a_{n}}=\frac{1-a_{n}^{2}}{2a_{n}}<0$\\
    $\therefore a_{n+1}<a_{n},\therefore{a_{n}}$单减\\
    $\therefore$由单调有界定理知$\lim_{n\to\infty}a_{n}$存在\\
    2)$\frac{a_{n}}{a_{n+1}}-1=\frac{a_{n}-a_{n+1}}{a_{n+1}}<a_{n-a_{n+1}}$\\
    $S_{n}=\sum_{n=1}^{n}a_{n}-a_{n+1}=a_{n}-a_{n+1}$\\
    $\because \lim_{n\to\infty}a_n$存在\\
    $\therefore {a_{n}}$一定有界,$\therefore {a_{1}-a_{n+1}}$有界\\
    $\therefore$由Th12.5知$\sum_{n=1}^{\infty}(\frac{a_{n}}{a_{n+1}}-1)$收敛
\end{solution}


\begin{problem}[本题满分15分]\\
    设函数$f$在$[a,+\infty)$有连续的导函数，且广义积分$\int_{a}^{+\infty}f(x)\ d x$和$\int_{u}^{+\infty}f^{\prime}(x)\ d x$都收敛，证明$\lim_{x\to\infty}f(x)=0$.
\end{problem}
\begin{solution}[本题15分]
   证明:\\
   设$A=\int_{a}^{+\infty}f^{\prime}(x)\ dx=\lim_{u\to+\infty}\int_{a}^{u}f^{\prime}(x)\ dx=\lim_{u\to+\infty}[f(u)-f(a)]$\\
   $\because \int_{u}^{+\infty}f^{\prime}(x)dx$收敛,$\therefore \lim_{u\to+\infty}f(u)$收敛\\
   $\therefore\lim_{u\to+\infty}f(u)=f(a)+A$\\
   又$\because \int_{a}^{+\infty}f(x)\ d x$收敛,设$\int_{a}^{+\infty}f(x)\ d x=B\neq0$,不妨设$A>0$\\
   $\therefore$取$\varepsilon=\frac{A}{2},\exists M>0,$当$x>M$时,$f(x)>\frac{B}{2}$\\
   $\therefore\int_{a}^{u}f(x)\ d x>\int_{a}^{u}\frac{B}{2}\ d x$\\
   $\therefore\lim_{u\to\infty}\int_{a}^{u}f(x)\ d x>\lim_{u\to\infty}\int_{a}^{u}\frac{B}{2}\ d x=+\infty $\\
   $\therefore\int_{a}^{+\infty}\frac{B}{2}\ d x $发散,\\
   $\therefore\int_{a}^{+\infty}f(x)\ d x$发散与$\int_{a}^{+\infty}f(x)\ d x$收敛矛盾\\
   $\therefore\lim_{x\to\infty}f(x)=f(a)+A=0$
\end{solution}


\begin{problem}[本题满分15分]\\
    已知$f(x,y)\doteq\begin{cases}\frac{x\dot{y}}{\sqrt{x^{2}+y^{2}}},&\text{若}(x,y)\neq(0,0)\\0,&\text{若}(x,y)=(0,0)\end{cases}$,证明$f(x,y)$在$(0,0)$连续，但不可微.
\end{problem}
\begin{solution}[本题15分]
   证明:\\
   $\lim_{(x,y)\to(0,0)}f(x,y)=\lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{x^{2}+y^{2}}}=0=f(0,0)$\\
   $\therefore f(x,y)$在$(0,0)$连续\\
   $f_{x}(0,0)=\lim_{\Delta x\to0}\frac{f(\Delta x,0)-f(0,0)}{\Delta x}=0$\\
   $f_{y}(0,0)=\lim_{\Delta y\rightarrow0}\frac{f(0,\Delta y)-f(0,0)}{\Delta y}=0$\\
   $\begin{aligned}
    &\lim_{(\rho\to0)}\frac{f(x,y)-f_x(0,0)x-f_y(0,0)y}{\rho}\\
    &=\lim_{\Delta x\to0,\Delta y\to0}\frac{\Delta x\cdot\Delta y}{\frac{\sqrt{\Delta x^{2}+\Delta y^{2}}}{\sqrt{\Delta x^{2}+\Delta y^{2}}}}\\
    &=\lim_{\Delta x\to0,\Delta y\to0}\frac{\Delta x\cdot\Delta y}{\Delta x^2+\Delta y^2}\\
    &=\lim_{\Delta x\to0,\Delta y=k\Delta x}\frac{k\Delta x^2}{\Delta x^2+k^2\Delta x^2}=\frac{k}{1+k^2}.
\end{aligned}$\\
与$k$有关,$\therefore$极限不存在\\
$\therefore$不可微
\end{solution}


\begin{problem}[本题满分15分]\\
    证明函数$f(x)=\sum_{n=1}^{\infty}\frac{\sin nx}{n^{2}+1}$在$(0,2\pi)$连续，且有连续的导数
\end{problem}
\begin{solution}[本题15分]
    证明:\\
    $u_{n}=|\frac{\sin nx}{n^{2}+1}|\leq\frac{1}{n^{2}+1}<\frac{1}{n^{2}}$\\
    $\sum_{n=1}^{\infty}\frac{1}{n^{2}}$收敛,$\therefore$由从判别法知$\sum_{n=1}^{\infty}\frac{\sin nx}{n^{2}+1}$一致收敛\\
    又每一项都连续,\\
    $\therefore f(x)=\sum_{n=1}^{\infty}\frac{\sin nx}{n^{2}+1}$在$(0,2\pi)$连续\\
    $f^{\prime}(x)=\sum_{n=1}^{\infty}\frac{n\cos nx}{n^{2}+1}$\\
    $|\frac{n\cos nx}{n^{2}+1}|\leq\frac{n}{n^{2}+1}=\frac{1}{n+\frac{1}{n}}$\\
    $\lim_{n\to\infty}\frac{1}{n+\frac{1}{n}}=0,\therefore\frac{n}{n^{2}+1}$一定有界\\
    $n>1,\{\frac{1}{n+\frac{1}{n}}\}$单减\\
    $\therefore$由单调有界定理知$\sum_{n=1}^{\infty}\frac{n}{n^{2}+1}$收敛\\
    $\therefore$由从判别法$\sum_{n=1}^{\infty}\frac{n\cos nx}{n^{2}+1}$一致收敛\\
    且每一项都连续,$\therefore f^{\prime}(x)$连续\\
    $\therefore f(x)=\sum_{n=1}^{\infty}\frac{\sin nx}{n^{2}+1}$有连续的导函数
 \end{solution}
\end{document}